\newproblem{lay:7_4_20}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.20}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if $A$ is a positive definite matrix, then an orthogonal diagonalization $A=PDP^T$ is a singular value decomposition of $A$.
}{
   % Solution
	Let us calculate $A^TA$ and consider the SVD $A=U\Sigma V^T$
	\begin{center}
		$A^TA=(PDP^T)^T(PDP^T)=PD^TP^TPDP^T=P(D^TD)P^T$
	\end{center}
	Since $A$ is positive definite $D^TD$ is a diagonal entry whose $ii$-th entry is $\lambda_i^2$. So its singular value is $\sigma_i=\sqrt{\lambda_i^2}=\lambda_i$.
	That is, for an SVD, we have $\Sigma=D$.
	
	By Exercise 7.4.19, we now that the columns of $V$ are the eigenvectors of $A^TA$. Given the diagonalization $A^TA=P(D^TD)P^T$ and the Diagonalization Theorem (Theorem 5.3.5),
	we see that the columns of $P$ are the eigenvectors of $A^TA$. So, we can make $V=P$
	
	Similarly, if we calculate $AA^T$ we have
	\begin{center}
		$AA^T=(PDP^T)(PDP^T)^T=PDP^TPD^TP^T=P(DD^T)P^T$
	\end{center}
	Again, this decomposition along with the Diagonalization Theorem show that the columns of $P$ are the eigenvectors of $AA^T$ and by Exercise 7.4.19, we can make $U=P$.
	
	Finally, the SVD decomposition of $A$ becomes
	\begin{center}
		$A=U\Sigma V^T=PDP^T$
	\end{center}
}
\useproblem{lay:7_4_20}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

